I found a statement in a book that I don't understand, it suggests that the number of DAGs on N nodes can be bounded from below by $$\prod \limits^N_{n=1} 2^{n-1} = 2^{N(N-1)/2}$$ My way of thinking is, if we have a DAG it should have maximum of N-1 edges. So $2^{N-1}$ should be the number of different subsets of N-1 edges (so it already includes all DAGs with N-2, N-3 edges).
However formula suggest that we still have to multiply this number by number of DAGs that has less edges than N-1, which I can't understand, since we already have all the N-2, N-3 subsets included in $2^{N-1}$.
Actually, a DAG with $N$ nodes can easily have more than $N-1$ edges: for example, let the nodes be $A,B,C$ and edges $A\rightarrow B, B\rightarrow C, A\rightarrow C$.
Fix an ordering of the nodes. Consider only DAGs such that all edges are directed so that that a parent precedes its child in the ordering. Now, the $n$th node has $n-1$ possible parents, thus the edges from nodes $(1,\ldots,n-1)$ can be picked in $2^{n-1}$ ways. The ordering ensures that there will be no cycles, and thus the parents of each node can be selected independently, and the total number of DAGs consistent with the ordering is the product \begin{equation} \prod_{n=1}^N 2^{n-1}. \end{equation}
However, there exist also DAGs that are not consistent with the selected ordering, and thus this number is only a lower bound for the total number of DAGs.